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The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42×10−4s−1 at a certain temperature. What is the half-life for this reaction? How long will it take for the concentration of SO2Cl2 to decrease to 25% of its initial concentration? If the initial concentration of SO2Cl2 is 1.00 M, how long will it take for the concentration to decrease to 0.78 M? If the initial concentration of SO2Cl2 is 0.150 M, what is the concentration of SO2Cl2 after 2.00×102s? After 5.00×102s?

Respuesta :

Answer:

a) Half life of the decomposition = 4951.1 s ≈ 4950 s

b) Time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900 s

c) If the initial concentration of SO₂Cl₂ is 1.00 M, time it will take for the concentration to decrease to 0.78 M is 1775s

d) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.146 M

e) If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10² s is 0.1398 M

Explanation:

Let C₀ represent the initial concentration of SO₂Cl₂

And C be the concentration of SO₂Cl₂ at anytime.

a) Rate of a first order reaction is represented by

dC/dt = - KC

dC/C = - kdt

Integrating the left hand side from C₀ to C₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)

In [(C₀/2)/C₀] = - k t(1/2)

In (1/2) = - k t(1/2)

- In 2 = - k t(1/2)

t₍₁,₂₎ = (In 2)/k

t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)

t₍₁,₂₎ = 4951.1 s

b) dC/C = - kdt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

In (C/C₀) = - kt

C/C₀ = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C = 25% of C₀ = 0.25C₀

0.25C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

- kt = In 0.25

- kt = - 1.386

t = 1.386/(1.4 × 10⁻⁴) = 9900 s

c) C = C₀ e⁻ᵏᵗ

C = 0.78 M; C₀ = 1.00 M

0.78 = 1 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.78

- kt = In 0.78

- kt = - 0.2485

t = 0.2485/(1.4 × 10⁻⁴) = 1775 s

d) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 2 × 10² s = 200 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972

C = 0.15 × 0.972 = 0.146 M

e) C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 5 × 10² s = 500 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324

C = 0.15 × 0.9324 = 0.1398 M

adefioyelaoye
  • The half life of the decomposition is 4951.1s.

  • The time it will take for the concentration of SO₂Cl₂ to decrease to 25% of its initial concentration = 9900s.

  • If the initial concentration of SO₂Cl₂ is 1.00 M, the time it will take for the concentration to decrease to 0.78 M is 1775s.

  • If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10²s is 0.146 M.

  • If the initial concentration of SO₂Cl₂ is 0.150 M, the concentration of SO₂Cl₂ after 2.00×10²s is 0.1398 M

Parameters

We can represent C₀ as the initial concentration of SO₂Cl₂.

C will be concentration of SO₂Cl₂ at anytime.

  • Rate of a first order reaction =

dC/dt = - KC

dC/C = - kdt

Integration will result in

In [(C₀/2)/C₀] = - k t(1/2)

In (1/2) = - k t(1/2)

- In 2 = - k t(1/2)

t₍₁,₂₎ = (In 2)/k

t₍₁,₂₎ = (In 2)/(1.4 × 10⁻⁴)

t₍₁,₂₎ = 4951.1 s

  • The time taken  is gotten by integrating  dC/C = - kdt

In (C/C₀) = - kt

C/C₀ = e⁻ᵏᵗ

C = C₀ e⁻ᵏᵗ

C = 25% of C₀ = 0.25C₀

0.25C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

- kt = In 0.25

- kt = - 1.386

t = 1.386/(1.4 × 10⁻⁴) = 9900 s

  • C = C₀ e⁻ᵏᵗ

C = 0.78 M; C₀ = 1.00 M

0.78 = 1 e⁻ᵏᵗ

e⁻ᵏᵗ = 0.78

- kt = In 0.78

- kt = - 0.2485

t = 0.2485/(1.4 × 10⁻⁴) = 1775 s

  • C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 2 × 10² s = 200 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 200) = 0.972

C = 0.15 × 0.972 = 0.146 M

  • C = C₀ e⁻ᵏᵗ

C₀ = 0.150 M, t = 5 × 10² s = 500 s

C = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = e^(-1.4 × 10⁻⁴ × 500) = 0.9324

C = 0.15 × 0.9324 = 0.1398 M

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